A certain number is given and the task is to count even digits and odd digits of the number and also even digits are present even a number of times and, similarly, for odd numbers.
Print Yes If: If number contains even digits even number of time Odd digits odd number of times Else Print No
Examples :
Input : 22233
Output : NO
count_even_digits = 3
count_odd_digits = 2
In this number even digits occur odd number of times and odd
digits occur even number of times so its print NO.
Input : 44555
Output : YES
count_even_digits = 2
count_odd_digits = 3
In this number even digits occur even number of times and odd
digits occur odd number of times so its print YES.
Efficient solution for calculating even and odd digits in a number.
C++
// C++ program to count // even and odd digits // in a given number #include <iostream> using namespace std; // Function to count digits int countEvenOdd(int n) { // Initialize event_count and odd_count int even_count = 0; int odd_count = 0; while (n > 0) { int rem = n % 10; // if condition is true then increment even_count if (rem % 2 == 0) { even_count++; } // increment odd_count else { odd_count++; } n = n / 10; } cout << "Even count : " << even_count; cout << "\nOdd count : " << odd_count; if (even_count % 2 == 0 && odd_count % 2 != 0) { return 1; } else { return 0; } } // Driver Code int main() { int n; n = 2335453; // Function call int t = countEvenOdd(n); if (t == 1){ cout << "\nYES" << endl; } else{ cout << "\nNO" << endl; } return 0; } |
Java
// Java program to count // even and odd digits // in a given number import java.io.*; class GFG { // Function to count digits static int countEvenOdd(int n) { // Initialize event_count and odd_count int even_count = 0; int odd_count = 0; while (n > 0) { int rem = n % 10; // if condition is true then increment even_count if (rem % 2 == 0){ even_count++; } // increment odd_count else { odd_count++; } n = n / 10; } System.out.println ( "Even count : " + even_count); System.out.println ( "Odd count : " + odd_count); if (even_count % 2 == 0 && odd_count % 2 != 0){ return 1; } else { return 0; } } // Driver Code public static void main (String[] args) { int n; n = 2335453; // Function call int t = countEvenOdd(n); if (t == 1) { System.out.println ( "YES" ); } else { System.out.println( "NO") ; } } } |
Python3
# python program to count even and # odd digits in a given number # Function to count digits def countEvenOdd(n): # Initialize event_count and odd_count even_count = 0 odd_count = 0 while (n > 0): rem = n % 10 # if condition is true then increment even_count if (rem % 2 == 0): even_count += 1 # increment odd count else: odd_count += 1 n = int(n / 10) print( "Even count : " , even_count) print("\nOdd count : " , odd_count) if (even_count % 2 == 0 and odd_count % 2 != 0): return 1 else: return 0 # Driver code n = 2335453; # Function call t = countEvenOdd(n); if (t == 1): print("YES") else: print("NO") # This code is contributed by Sam007. |
C#
// C# program to count even and // odd digits in a given number using System; class GFG { // Function to count digits static int countEvenOdd(int n) { // Initialize event_count and odd_count int even_count = 0; int odd_count = 0; while (n > 0) { int rem = n % 10; // if condition is true then increment even_count if (rem % 2 == 0) { even_count++; } else { odd_count++; } n = n / 10; } Console.WriteLine("Even count : " + even_count); Console.WriteLine("Odd count : " + odd_count); if (even_count % 2 == 0 && odd_count % 2 != 0) { return 1; } else { return 0; } } // Driver Code public static void Main () { int n; n = 2335453; // Function call int t = countEvenOdd(n); if (t == 1) { Console.WriteLine ("YES"); } else { Console.WriteLine("NO") ; } } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to count // even and odd digits // in a given number // Function to count digits function countEvenOdd($n) { // Initialize event_count and odd_count $even_count = 0; $odd_count = 0; while ($n > 0) { $rem = $n % 10; // if condition is true then increment even_count if ($rem % 2 == 0) { $even_count++; } // increment odd_count else { $odd_count++; } $n = (int)($n / 10); } echo("Even count : " . $even_count); echo("\nOdd count : " . $odd_count); if ($even_count % 2 == 0 && $odd_count % 2 != 0) { return 1; } else { return 0; } } // Driver code $n = 2335453; // Function call $t = countEvenOdd($n); if ($t == 1) { echo("\nYES"); } else { echo("\nNO"); } // This code is contributed by Ajit. ?> |
Javascript
<script> /// Javascript program to count // even and odd digits // in a given number // Function to count digits function countEvenOdd(n) { let even_count = 0; let odd_count = 0; while (n > 0) { let rem = n % 10; // if condition is true then increment even_count if (rem % 2 == 0) { even_count++; } // increment odd_count else { odd_count++; } n = Math.floor(n / 10); } document.write("Even count : " + even_count); document.write("<br>" + "Odd count : " + odd_count); if (even_count % 2 == 0 && odd_count % 2 != 0) { return 1; } else { return 0; } } // Driver Code let n; n = 2335453; // Function call let t = countEvenOdd(n); if (t == 1) { document.write("<br>" + "YES" + "<br>"); } else { document.write("<br>"+"NO" + "<br>"); } // This code is contributed by Mayank Tyagi </script> |
Output
Even count : 2 Odd count : 5 YES
Time Complexity: O(n)
Auxiliary Space: O(1)
Another solution to solve this problem is a character array or string.
C++
// C++ program to count // even and odd digits // in a given number // using char array #include <bits/stdc++.h> using namespace std; // Function to count digits int countEvenOdd(char num[], int n) { int even_count = 0; int odd_count = 0; for (int i = 0; i < n; i++) { int x = num[i] - 48; if (x % 2 == 0) { even_count++; } else { odd_count++; } } cout << "Even count : " << even_count; cout << "\nOdd count : " << odd_count; if (even_count % 2 == 0 && odd_count % 2 != 0) { return 1; } else { return 0; } } // Driver Code int main() { char num[18] = { 1, 2, 3 }; int n = strlen(num); // Function cal int t = countEvenOdd(num, n); if (t == 1) { cout << "\nYES" << endl; } else { cout << "\nNO" << endl; } return 0; } |
Java
// Java program to count // even and odd digits // in a given number // using char array import java.io.*; class GFG { // Function to count digits static int countEvenOdd(char num[], int n) { int even_count = 0; int odd_count = 0; for (int i = 0; i < n; i++) { int x = num[i] - 48; if (x % 2 == 0) { even_count++; } else { odd_count++; } } System.out.println ("Even count : " + even_count); System.out.println( "Odd count : " + odd_count); if (even_count % 2 == 0 && odd_count % 2 != 0) { return 1; } else { return 0; } } // Driver Code public static void main (String[] args) { char num[] = { 1, 2, 3 }; int n = num.length; // Function call int t = countEvenOdd(num, n); if (t == 1) { System.out.println("YES") ; } else { System.out.println("NO") ; } } } // This code is contributed by vt_m |
Python3
# Python3 program to count # even and odd digits # in a given number # using char array # Function to count digits def countEvenOdd(num, n): even_count = 0; odd_count = 0; num=list(str(num)) for i in num: if i in ('0','2','4','6','8'): even_count+=1 else: odd_count+=1 print("Even count : ", even_count); print("Odd count : ", odd_count); if (even_count % 2 == 0 and odd_count % 2 != 0): return 1; else: return 0; # Driver Code num = (1, 2, 3); n = len(num); t = countEvenOdd(num, n); if t == 1: print("YES"); else: print("NO"); # This code is contributed by mits. |
C#
// C# program to count // even and odd digits // in a given number // using char array using System; class GFG { // Function to count digits static int countEvenOdd(char []num, int n) { int even_count = 0; int odd_count = 0; for (int i = 0; i < n; i++) { int x = num[i] - 48; if (x % 2 == 0) even_count++; else odd_count++; } Console.WriteLine("Even count : " + even_count); Console.WriteLine( "Odd count : " + odd_count); if (even_count % 2 == 0 && odd_count % 2 != 0) return 1; else return 0; } // Driver code public static void Main () { char [] num = { '1', '2', '3' }; int n = num.Length; int t = countEvenOdd(num, n); if (t == 1) Console.WriteLine("YES") ; else Console.WriteLine("NO") ; } } // This code is contributed by Sam007. |
PHP
<?php // PHP program to count // even and odd digits // in a given number // using char array // Function to count digits function countEvenOdd($num, $n) { $even_count = 0; $odd_count = 0; for ($i = 0; $i < $n; $i++) { $x = $num[$i] - 48; if ($x % 2 == 0) $even_count++; else $odd_count++; } echo "Even count : " . $even_count; echo "\nOdd count : " . $odd_count; if ($even_count % 2 == 0 && $odd_count % 2 != 0) return 1; else return 0; } // Driver Code $num = array( 1, 2, 3 ); $n = strlen(num); $t = countEvenOdd($num, $n); if ($t == 1) echo "\nYES" ,"\n"; else echo "\nNO" ,"\n"; // This code is contributed by ajit. ?> |
Javascript
<script> // Javascript program to count // even and odd digits // in a given number // using char array // Function to count digits function countEvenOdd(num, n) { even_count = 0; odd_count = 0; for(var i = 0; i < n; i++) { x = num[i] - 48; if (x % 2 == 0) even_count++; else odd_count++; } document.write("Even count : " + even_count); document.write("<br>Odd count : " + odd_count); if (even_count % 2 == 0 && odd_count % 2 != 0) return 1; else return 0; } // Driver code var num = [ 1, 2, 3 ]; n = num.length; t = countEvenOdd(num, n); if (t == 1) document.write("<br>YES <br>"); else document.write("<br>NO <br>"); // This code is contributed by akshitsaxenaa09 </script> |
Output
Even count : 1 Odd count : 2 NO
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #3:Using typecasting(Simplified Approach):
- We have to convert the given number to a string by taking a new variable.
- Traverse the string, convert each element to an integer.
- If the character(digit) is even, then the increased count
- Else increment the odd count.
- If the even count is even and the odd count is odd, then print Yes.
- Else print no.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach #include<bits/stdc++.h> using namespace std; string getResult(int n) { // Converting integer to String string st = to_string(n); int even_count = 0; int odd_count = 0; // Looping till length of String for(int i = 0; i < st.length(); i++) { if ((st[i] % 2) == 0) // Digit is even so increment even count even_count += 1; else odd_count += 1; } // Checking even count is even and // odd count is odd if (even_count % 2 == 0 && odd_count % 2 != 0) return "Yes"; else return "no"; } // Driver Code int main(){ int n = 77788; // Passing this number to get result function cout<<getResult(n)<<endl; } // This code is contributed by shinjanpatra. |
Java
// Java implementation of above approach class GFG{ static String getResult(int n) { // Converting integer to String String st = String.valueOf(n); int even_count = 0; int odd_count = 0; // Looping till length of String for(int i = 0; i < st.length(); i++) { if ((st.charAt(i) % 2) == 0) // Digit is even so increment even count even_count += 1; else odd_count += 1; } // Checking even count is even and // odd count is odd if (even_count % 2 == 0 && odd_count % 2 != 0) return "Yes"; else return "no"; } // Driver Code public static void main(String[] args) { int n = 77788; // Passing this number to get result function System.out.println(getResult(n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python implementation of above approach def getResult(n): # Converting integer to string st = str(n) even_count = 0 odd_count = 0 # Looping till length of string for i in range(len(st)): if((int(st[i]) % 2) == 0): # digit is even so increment even count even_count += 1 else: odd_count += 1 # Checking even count is even and odd count is odd if(even_count % 2 == 0 and odd_count % 2 != 0): return 'Yes' else: return 'no' # Driver Code n = 77788 # passing this number to get result function print(getResult(n)) # this code is contributed by vikkycirus |
C#
// C# implementation of above approach using System; using System.Collections.Generic; class GFG{ static String getResult(int n) { // Converting integer to String String st = String.Join("",n); int even_count = 0; int odd_count = 0; // Looping till length of String for(int i = 0; i < st.Length; i++) { if ((st[i] % 2) == 0) // Digit is even so increment even count even_count += 1; else odd_count += 1; } // Checking even count is even and // odd count is odd if (even_count % 2 == 0 && odd_count % 2 != 0) return "Yes"; else return "no"; } // Driver Code public static void Main(String[] args) { int n = 77788; // Passing this number to get result function Console.WriteLine(getResult(n)); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript implementation of above approach function getResult(n) { // Converting integer to String var st = n.toString(); var even_count = 0; var odd_count = 0; // Looping till length of String for(var i = 0; i < st.length; i++) { if ((st.charAt(i) % 2) == 0) // Digit is even so increment even count even_count += 1; else odd_count += 1; } // Checking even count is even and // odd count is odd if (even_count % 2 == 0 && odd_count % 2 != 0) return "Yes"; else return "no"; } // Driver Code var n = 77788; // Passing this number to get result function document.write(getResult(n)); </script> |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(1)
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